Answer:
The % yield is 90.68 %
Step-by-step explanation:
Step 1: Data given
Mass of acetic acid = 6.103 grams
Mass of isoamyl alcohol = 3.728 grams
Mass isoamyl acetate obtained = 4.993 grams
Molar mass acetic acid = 60.05 g/mol
Molar mass of isoamyl alcohol = 88.15 g/mol
Molar mass of isoamyl acetate = 130.19 g/mol
Step 2: The balanced equation
C5H12O + CH3COOH → C7H14O2 +H2O Isoamyl alcohol + acetic acid ---> Isoamyl acetate + water
Step 3: Calculate moles of acetic acid
Moles acetic acid = Mass acetic acid / Molar mass acetic acid
Moles acetic acid = 6.103 grams / 60.05 g/mol
Moles acetic acid = 0.1016 moles
Step 4: Calculate moles isoamyl alcohol
Moles isoamyl alcohol = 3.728 grams / 88.15 g/mol
Moles isoamy lalcohol = 0.04229 moles
Step 5: Calculate limiting reactant
The mole ratio is 1:1 so isoamyl alcohol has the smallest number of moles. So isoamyl alcohol is the limiting reactant. It will be completely be consumed (0.04229 moles).
Acetic acid is in excess, there will be consumed 0.04229 moles.
There will remain 0.1016 - 0.04229 = 0.05931 moles
Step 6: Calculate moles of isoamyl acetate
For 1 mole acetic acid , we need 1 mole of isoamyl alcohol, to produce 1 mole isoamyl acetate and 1 mole of H2O
For 0.04229 moles isoamyl alcohol consumed, there will be produced 0.04229 moles of isoamyl acetate.
Step 7: Calculate mass of isoamyl acetate
Mass isoamyl acetate = moles isoamyl acetate * molar mas isoamyl acetate
Mass isoamyl acetate = 0.04229 moles * 130.19 g/mol
Mass isoamyl acetate = 5.506 grams = theoretical yield
Step 8: Calcuate % yield
% yield = actual yield / theoretical yield
% yield = (4.993 / 5.506)*100%
% yield = 90.68 %
The % yield is 90.68 %